Question

pre cal. will give brainiest

Answer 1

Answer:

D. $-\dfrac{1}{2}$

Step-by-step explanation:

Given the fraction:

$\dfrac{\dfrac{3}{4y}-\dfrac{2}{y}}{\dfrac{1}{y}+\dfrac{3}{2y}}$

Consider the numerator and the denominator separately:

Numerator:

$\dfrac{3}{4y}-\dfrac{2}{y}=\dfrac{3}{4y}-\dfrac{8}{4y}=\dfrac{3-8}{4y}=-\dfrac{5}{4y}$

Denominator:

$\dfrac{1}{y}+\dfrac{3}{2y}=\dfrac{2}{2y}+\dfrac{3}{2y}=\dfrac{2+3}{2y}=\dfrac{5}{2y}$

Now,

$\dfrac{\dfrac{3}{4y}-\dfrac{2}{y}}{\dfrac{1}{y}+\dfrac{3}{2y}}=\dfrac{-\dfrac{5}{4y}}{\dfrac{5}{2y}}=-\dfrac{5}{4y}\div \dfrac{5}{2y}=-\dfrac{5}{4y}\times \dfrac{2y}{5}=-\dfrac{1}{2}$